Reverse laplace transformation involving the laplace of unit step function multiplied by another function

Question

I am confused how to solve this function, I want the laplace transformation of it:

$\frac{2e^{-s}}{(s+1)^3}$

These are relevant rules I know:

$ \mathcal{L}^{-1}[F(s)] = f(t) => \mathcal{L}^{-1}[e^{-sa} F(s)] = u(t-a)f(t-a) $

Is this correct?

And this rule:

$\mathcal{L} [u^{(n)}(t) \cdot g(t)] = e^{-ns} \mathcal{L}[g(t+n)]$

Is this correct?

Can they both be correct? The point that confuses me is that in the second rule the exponential is out of the laplace, so a laplace transformation of it would be:

$\mathcal{L} [u^{(n)}(t) \cdot g(t)] = \mathcal{L}[ e^{-ns} \mathcal{L}[g(t+n)]]$

1. which one of the 2 rules is correct?

2. What is the rule about taking funtions out of the laplace? (I thought you can only do that to constants)

3. How does convolution play a part here? I know that multiplication in one area means convolution in the other area but these rules aren't about convolution right?

4. How do I apply the rules on the function?

Tnx

Edit: as no one answered yet I'll add the following:

I try to use Laplace's displacement theorem and get this:

$ \mathcal{L}[2e^{-s} \cdot \frac{1}{(s+1)^3}] = 2 \mathcal{L}[ \frac{1}{(s+1)^3}] (t+1) = (t+1)^2e^{-(t+1)} $

Did I get it right? this isn't what symbolab gives and I don't understand the mistake if there is one. Symbolab's answer is:

$ H(t-1)e^{-t+1}(t-1)^2 $